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Calculate the mass of benzoic acid required to prepare 100cm3 each of 0.1m,0.25m, and 0.5m solutions

User Icon for bobpursley bobpursley answered
7 years ago

I assume your m should be Molarity, M.

mass=molartiy*molmass*volumeinLiters
mass=.1*molmassBenzoicAcid*.1m^3
so the mom mass is 122g/mole, so we have
mass=.1 Mole/liter*122g/mole*.1 Liter= 1.22 grams
check my thinking.

User Icon for Benjamin's Benjamin's answered
3 years ago

M =molar.volume. molar mass÷1000

314×220×1.28÷ 1000
= 0.535

User Icon for Explain Bot Explain Bot answered
11 months ago

To calculate the mass of benzoic acid required to prepare the given solutions, we need to use the formula:

Mass (g) = Molarity (mol/L) x Volume (L) x Molar mass (g/mol)

First, let's calculate the amount of benzoic acid required to prepare a 0.1 M solution.

Molarity (mol/L) = 0.1 mol/L
Volume (L) = 100 cm^3 ÷ 1000 = 0.1 L (since 1 cm^3 = 0.001 L)
Molar mass of benzoic acid (C7H6O2) = 122.12 g/mol

Mass (g) = 0.1 mol/L x 0.1 L x 122.12 g/mol
Mass = 1.2212 g

Therefore, to prepare a 0.1 M solution, you would need 1.2212 grams of benzoic acid.

Now let's move on to the 0.25 M solution.

Molarity (mol/L) = 0.25 mol/L
Volume (L) = 100 cm^3 ÷ 1000 = 0.1 L
Molar mass of benzoic acid (C7H6O2) = 122.12 g/mol

Mass (g) = 0.25 mol/L x 0.1 L x 122.12 g/mol
Mass = 3.052 g

Therefore, to prepare a 0.25 M solution, you would need 3.052 grams of benzoic acid.

Lastly, let's calculate the mass of benzoic acid required for a 0.5 M solution.

Molarity (mol/L) = 0.5 mol/L
Volume (L) = 100 cm^3 ÷ 1000 = 0.1 L
Molar mass of benzoic acid (C7H6O2) = 122.12 g/mol

Mass (g) = 0.5 mol/L x 0.1 L x 122.12 g/mol
Mass = 6.106 g

Therefore, to prepare a 0.5 M solution, you would need 6.106 grams of benzoic acid.